# 14.E: Homework Chapter 14 Answers

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Dilutions with Strong Acids and Strong Bases:

1. V1 = 350. mL of the strong acid will be needed to neutralize the strong base!

3. V1 = 48.0 mL of the strong acid will be needed to neutralize the strong base!

Neutralization Reactions:

5.

a.) NaOH(aq) + HI(aq) → H2O(l) + NaI(aq)

b.) KOH(aq) + HCl(aq) → H2O(l) + KCl(aq)

c.) RbOH(aq) + HBr(aq) → H2O(l) + RbBr(aq)

d.) Ba(OH)2(aq) + 2HClO4(aq) → 2H2O(l) + Ba(ClO4)2(aq)

7. For the following pairs, write their neutralization reaction.

a.) Ca(OH)2(aq) + 2HCl(aq) → 2H2O(l) + CaCl2(aq)

b.) 2NaOH(aq) + H2SO4(aq) → 2H2O(l) + Na2SO4(aq)

c.) Ba(OH)2(aq) + 2HNO3(aq) → 2H2O(l) + Ba(NO3)2(aq)

d.) CsOH(aq) + HBr(aq) → H2O(l) + CsBr(aq)

9.

a.) Ca(s) + 2HCl(aq) → H2(g) + CaCl2(aq)

b.) 2Li(s) + 2HCl(aq) → H2(g) + 2LiCl(aq)

c.) 2K(s) + 2HCl(aq) → H2(g) + 2KCl(aq)

d.) Ba(s) + 2HCl(aq) → H2(g) + BaCl2(aq)

Strong Acids and Bases vs. Weak Acids and Bases:

11. a.) WEAK   b.) STRONG   c.) WEAK   d.) STRONG

13.   a.) STRONG   b.) STRONG   c.) WEAK   c.) WEAK

15.

a.) [H3O+] = less than 12.0 M

b.) [H3O+] = 0.010 M

c.) [H3O+] = less than 1.50 M

d.) [H3O+] = 4.30 M

17.

a.) [OH-] = 1.00 M

b.) [OH-] = less than 4.50 M

c.) [OH-] = 1.50 M

d.) [OH-] = 2.30 M

19.

a.) [OH-] = 1.00 M

b.) [OH-] = 3.06 M

c.) [OH-] = 14.0 M

d.) [OH-] = less than 0.00000010 M

BrØnsted-Lowry/Arrhenius Acids and Bases:

21. The base (LiOH) acts as an Arrhenius base because OH- was produced.

23. The base (NH3) is acting as a BrØnsted-Lowry base because it accepted a proton. It started as NH3, and then became NH4+.

25.

a.) Acid,  H2SO4(aq) → H+(aq) + HSO4-(aq)

b.) Base, NaOH(aq) → Na+(aq) + OH-(aq)

c.) Acid, CH3COOH(aq) → H+(aq) + CH3COO-(aq)

d.) Base, Ca(OH)2(aq) → Ca2+(aq) + 2OH-(aq)

27.

a.) Acid, HCl(aq) → H+(aq) + Cl-(aq)

b.) Acid, HClO4(aq) → H+(aq) + ClO4-(aq)

c.) Acid, CH2O2(aq) → H+(aq) + CHO2-(aq)

d.) Base, KOH(aq) → K+(aq) + OH-(aq)

BrØnsted-Lowry Acids/Bases & Conjugate Acids/Bases: 29.

a.) HCN: B-Acid; H2O: B-Base

b.) NH3: B-Base; H2O: B-Acid

c.) F-: B-Bases; H2O: B-Acid

d.) H2CO3: B-Acid; H2O: B-Base

31.

a) HF: B-Acid; OH-: B-Base; F-: C-Base; H2O: C-Acid

b.) H3PO4: B-Acid; OH-: B-Base; H2PO4-: C-Base; H2O: C-Acid

c.) HCl: B-Acid; H2O: B-Base; H3O+: C-Acid; Cl-: C-Base

d.) CH3COOH: B-Acid; H2O: C-Base; H3O+: C-Acid; CH3COO-: C-Base

33.

a.) Yes, this is a conjugate acid-base pair

b.) No, this is not a conjugate acid-base pair

c.) Yes, this is a conjugate acid-base pair

d.) Yes, this is a conjugate acid-base pair

35. For the following pairs, state whether they are or not conjugate acid-base pairs.

a.) Yes, this is a conjugate acid-base pair

b.) No, this is not a conjugate acid-base pair

c.) Yes, this is a conjugate acid-base pair

d.) No, this is not a conjugate acid-base pair

37.  a.)    CNO-   b.)   F-   c.) HS-   d.) IO3-

39.   a.) HCO3-   b.) H3PO4   c.) HSO4-   d.) HClO4

41.    a.) H2C2O4   b.) HPO42-   c.) CH3COOH   d.) H2C2COOH

pH & pOH Calculations/Acidic vs. Basic Solutions:

43. a.) Basic   b.) Acidic   c.) Neutral   d.) Basic

45. a.) Basic   b.) Acidic   c.) Basic   d.) Acidic 47.

a.) 1.0 x10-6 M

b.) 4.0 x10-5 M

c.) 6.7 x10-3 M

d.) 1.0 x10-9 M 49. A pH of 7 is a neutral solution, which means that there is the same amount of acid and base in the solution.

51. 8   9   10   11   12   13   14

53. a.) Acidic    b.) Acidic    c.)  Neutral     d.) Basic 55. a.) 10.00    b.) 7.57    c.) 5.00   d.) 8.52 57. a.) 5.00    b.) 1.46   c.) 3.70   d.) 4.40 59. a.) 9.82    b.) 6.46    c.) 14.10    d.) 3.70 61. a.) 0.032 M    b.) 1.0 x10-4 M   c.) 3.2 x10-7 M   d.) 1.3 x10-4 M

63. From the following pH values, determine the hydronium concentration.

a.) 1.5 x10-6 M    b.) 6.2 x10-5 M   c.) 0.66 M   d.) 0.010 M 65. a.) 1.0 x10-11 M   b.) 4.8 x10-10 M   c.) 1.0 x10-8 M   d.) 1.0 x10-14 M 67.

a.)  pOH = -log (1.0 x10-8) = 8.00       pH = 14.00 – 8.00 = 6.00

b.) pOH = -log (2.5 x10-10) = 9.60      pH = 14.00 – 9.60 = 4.40

c.) pOH = -log (4.0 x10-9) = 8.40        pH = 14.00 – 8.40 = 5.60

d.) pOH = -log (3.8 x10-5) = 4.42         pH = 14.00 – 4.42 = 9.58

69. a.) pH = -log (0.0010) = 3.00

b.) pOH = -log (0.020) = 1.70    ;    pH = 14.00 – 1.70 = 12.30

c.) pH = -log (0.0030) = 2.52

d.) pH = -log (0.0040) = 2.40

71. a.) pH = -log (0.00321) = 2.49

b.) pOH = -log (0.0193) = 1.71    ;     pH = 14.00 – 1.71 = 12.29

c.) pH = -log (0.000483) = 3.32

d.) pOH = -log (0.148) = 0.83       ;     pH = 14.00 – 0.83 = 13.17

73. pH = 14.00 – 6.30 = 7.70 ; Basic

b.) pH = 14.00 – 3.98 = 10.02 ; Basic

c.)  pH = 14.00 – 12.74 = 1.26 ; Acidic

d.)  pH = 14.00 – 8.29 = 5.71 ; Acidic

75. d.) pH = 12.83

76. a.) pH = 1.89

77. c.) pOH = 1.54

Cumulative/Challenge Problems:

79. [NaOH] = 0.0300 M

The concentration of the strong base is greater than the concentration of the strong acid, which indicates that the solution may be basic.

81. pH = -log (1.0 x10-11) = 11.00

pOH = 14.00 – 11.00 = 3.00 ; The solution is basic.

83. According to the BrØnsted-Lowry definitions of acids and bases, acids are proton (hydrogen ion) donors while bases are proton acceptors. With this information, we know that all acids must have a hydrogen ion to give away, while all bases only have to be able to accept a hydrogen ion. They do not necessarily have to have a hydroxide ion to do this. If the base does have a hydroxide ion, then the hydrogen ion accepted will form water with the hydroxide ion present.

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